Difference between revisions of "1971 Canadian MO Problems/Problem 2"
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== Solution == | == Solution == | ||
+ | ===Solution 1=== | ||
<cmath>\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right) \ge 9</cmath> | <cmath>\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right) \ge 9</cmath> | ||
<cmath>1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{xy} \ge 9 </cmath> | <cmath>1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{xy} \ge 9 </cmath> | ||
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which is true since by AM-GM, we get: <cmath>x^2 + y^2 \ge 2xy</cmath> <cmath>x^2 + 2xy + y^2 \ge 4xy</cmath> <cmath> (x+y)^2 \ge 4xy</cmath> and we are given that <math> x + y = 1</math>, so <cmath> (x+y)^2 \ge 4xy \Rightarrow 1 \ge 4xy </cmath> | which is true since by AM-GM, we get: <cmath>x^2 + y^2 \ge 2xy</cmath> <cmath>x^2 + 2xy + y^2 \ge 4xy</cmath> <cmath> (x+y)^2 \ge 4xy</cmath> and we are given that <math> x + y = 1</math>, so <cmath> (x+y)^2 \ge 4xy \Rightarrow 1 \ge 4xy </cmath> | ||
+ | ===Solution 2=== | ||
+ | Let <math>f(x)=\left(1+\frac{1}{x}\right)\left(1+\frac{1}{1-x}\right)</math>. Since we want to find the minimum of the function over the [[interval]] <math>(0,1)</math>, we can take the [[derivative]]. Using the [[product rule]], we get <math>\frac{2(2x-1)}{x^2(1-x)^2}</math>. Since we want this value to be zero, the numerator must be zero. But the only root of the equation <math>2(2x-1)</math> is <math>1/2</math>, and so plugging the answer back in we have | ||
+ | <cmath>(1+\frac{1}{\frac{1}{2}})(1+\frac{1}{\frac{1}{2}})</cmath> | ||
+ | <cmath>(1+2)(1+2)</cmath> | ||
+ | <cmath>3\cdot 3</cmath> | ||
+ | <cmath>9</cmath> | ||
+ | Thus the minimum is <math>\boxed{9}</math> | ||
== See Also == | == See Also == | ||
{{Old CanadaMO box|num-b=1|num-a=3|year=1971}} | {{Old CanadaMO box|num-b=1|num-a=3|year=1971}} | ||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] |
Revision as of 17:31, 16 January 2023
Problem
Let and
be positive real numbers such that
. Show that
.
Solution
Solution 1
which is true since by AM-GM, we get:
and we are given that
, so
Solution 2
Let . Since we want to find the minimum of the function over the interval
, we can take the derivative. Using the product rule, we get
. Since we want this value to be zero, the numerator must be zero. But the only root of the equation
is
, and so plugging the answer back in we have
Thus the minimum is
See Also
1971 Canadian MO (Problems) | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • | Followed by Problem 3 |